Gravity is a force because it pulls down on objects.
Answer:
16.8ohms
Explanation:
According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.
Mathematically, V = IRt where;
V is the voltage across the circuit
I is the current
R is the effective resistance
For a series connected circuit, same current but different voltage flows through the resistors.
If the initial current in a circuit is 19.3A,
V = 19.3R... (1)
When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;
V = 13.4(7.4+R)... (2)
Note that the initial resistance is added to the additional resistance because they are connected in series.
Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;
19.3R = 13.4(7.4+R)
19.3R = 99.16+13.4R
19.3R-13.4R = 99.16
5.9R = 99.16
R= 99.16/5.9
R = 16.8ohms
The resistance in the original circuit will be 16.8ohms
It depends on how you want to solve it you can solve it in many different meathods:$
Answer: work must be done on the system (Option A)
Explanation:
The second law of thermodynamics is the fundamental law of nature; it states that energy can be transferred from cold objects to hot objects only, if work is done on the system. If energy is added to the system then as a result the thermal energy would increase. Second law of thermodynamics is used to determine whether a process is spontaneous or not. Moreover,the second law of thermodynamics is also used in refrigerators.
Answer:

Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.

Now, substitute this into 'dF':

Now we can integrate dF over the rod.
